Problem with differential form of Maxwell's third equation?

[lalbatros]

understanding electromagnetic water flow meters

Hello,

I posted the following a few weeks ago:

https://www.physicsforums.com/showthread.php?t=118872"

Actually, asking this is equivalent to ask:

how can a motional field be measured with a voltmeter,
and in the flowmeter, is the motional field really measured,
and is a voltmeter measuring and electric field,
and if yes does it mean that a motional field ...​

Any insight ?

Michel

 

[lalbatros]

is it a coïncidence ...

Let's have a look at the well known identity:

\frac{d}{dt} \int_{St} B.dS = \int_{St} \frac{\partial B}{\partial t}.dS + \int_{\partial St}v \times B .ds

The vXB term that you see there, this is just the motional field.
Is it a coïncidence that it is precisely this term that appears in the Lorentz force ?
And finally, why is called 'motional field' ?

 

[leright]

Imagine that in the frame (A) where the charge is stationary, the E field is 0, and the B field is constant. The lorentz force is 0 in this case.
Now, go to a frame B which is moving with velocity V wrt A. In this frame (B), there is an electric and a magnetic field: E = V x B and B remains what it was. Now, the total Lorentz force in this frame, on the charge, which is moving with velocity -V in this frame, is given by:
F = q (E + (-V) x B) = q ( V x B + (-V) x B) = 0 too.

It must be late, because I don't see how you got to the conclusion that E = V x B in the non-stationary reference frame. I see how you got the magnetic force though...

 

[leright]

Let's have a look at the well known identity:

\frac{d}{dt} \int_{St} B.dS = \int_{St} \frac{\partial B}{\partial t}.dS + \int_{\partial St}v \times B .ds

The vXB term that you see there, this is just the motional field.
Is it a coïncidence that it is precisely this term that appears in the Lorentz force ?
And finally, why is called 'motional field' ?

That's exactly what I've been bashing my head against the wall for days about. When Maxwell originally formulated the original equations he made the assumption that motional EMF was not occurring and the conditions were free space and when he commuted the derivative into the integral he only brought in a partial derivative. This inherently leaves a part of the EMF out of the picture if you define EMF as a force integrated around a closed path (motional EMF), but it is compatible with the Lotentz force equation commonly reffered to in textbooks. Also, as Vanesch pointed out, if an E-field is considered to be the cause of the force due to a moving charge in a b-field then 3 different charges moving at three different velocities pass a common point in space, they will all have different forces, or in other words they will all have differing field desciptions. This makes it difficult to deal with large groups of charge with different charges moving at varying velocities (like in a plasma), because each charge will have its own unique field description. A field should be a description of the means by which various particles are acted upon in space, regardless of the size, orientation, or velocity of said particles. Maxwell's E-fields allow such a consistent description. In Maxwell's formulation, the E-field is the same for all charges. Also, the B-field is the same for all moving charges. However, in Hertz's formulation, the E-field for each particle differs and depends on the respective velocities of each particle.

What I am curious about though, is if there is a charge moving in a uniform B-field with a force acting upon it, and this charge passes by another stationary charge, will the additional stationary charge feel the force set up by the B-field and the other moving charge, in addition to the force the stationary charge feels by the moving charge itself (due to the coulomb force). In addition, in what direction does this occur.

OK, I am getting exhausted and I need to get up at 6am tomorrow for work. peace.

 

[vanesch]

It must be late, because I don't see how you got to the conclusion that E = V x B in the non-stationary reference frame. I see how you got the magnetic force though...

The transformation rules of the (E,B) field from one coordinate frame to another are given by:

http://farside.ph.utexas.edu/teaching/jk1/lectures/node24.html

In the non-relativistic limit, this comes down to:

If (E,B) is the field in a coordinate frame 1, and (E',B') is the field in a coordinate frame 2 which has velocity V wrt the coordinate frame 1, then this reduces to:

E' = E + V x B
B' = B

This must be the case of course, if the Lorentz force is to be consistent:

If you want to have the same force to act in the two reference frames, you have to have, for a particle which has velocity v in frame 1 (and hence, non-relativistically, v - V in frame 2):

F = q (E + v x B) = q (E' + (v - V) x B')

for all possible velocities v, then you quickly find the above-cited transformation rules.This is what I meant with "the E and B fields do not transform as vectors under boosts" in my post #3 in this thread...cheers,
Patrick.

 

[creat]

Well, the first equation is already wrong, so I stopped reading. You cannot permute the integral and the d/dt if the surface is moving.

This is easily verified: take the vector field a to be equal to (3,0,0) when (x,y,z) is within the cube (0<x<1, 0<y<1, 0<z<1) and consider a unit square parallel to the yz plane, which is in the plane x = 0.5 and which touches the Y and Z axis at t = 0, and is moving with a velocity = 5 in the z-direction.
Clearly, d a / dt = 0 because it is a static field, independent of t, so the right hand is 0. However, the left hand isn't: the flux over this moving surface is 3 at t=0, and decreases linearly to 0 at t = 0.2, so d phi/dt = -15.

I am the author of the site. In the meantime I have updated the site:
http://www.angelfire.com/sc3/elmag
and the page:
http://www.angelfire.com/sc3/elmag/files/EM05FL.pdf

You are wrong with this example. The field on the unit square is not static, for
a = (3,0,0), for v*t <= z < 1
a = (0,0,0), for 1 <= z <= 1+v*t.
So d a / dt = -15 delta(v*t), (delta is Dirac's delta function)
After integration we have that left side is equal to the right.
Nobody's perfect, except the Heavenly Father